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Hello students, in this question we have to discuss the stoichiometry for oxidation of butane.
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2 moles of butane combined with 13 moles of oxygen to produce 8 moles of carbon dioxide and 10 moles of water.
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In the first case, the mass of butane has been given to us as 9 .03 grams and we have to find out the mass of water produced.
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The number of moles of butane would be equal to its given mass divided by its molar weight.
00:43
The given mass is 9 .03 gram and its molar weight is 58 .12 grams per mole.
00:52
The number of moles of butane come out to be 0 .15 mole.
00:57
2 mole of butane produce 10 moles of water.
01:08
0 .15 mole of butane would produce 10 divided by 2 multiplied by 0 .15 mole of water.
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This comes out to be 0 .75 mole.
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The mass of water produced would be equal to its number of moles multiplied by its molar weight.
01:37
The number of moles are 0 .75 and molar weight of water is 18 grams per mole.
01:44
The mass of water formed comes out to be 13 .5 grams...